Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The following game is played. The player randomly draws from the set of integers \(\\{1,2, \ldots, 20\\} .\) Let \(x\) denote the number drawn. Next the player draws at random from the set \(\\{x, \ldots, 25\\}\). If on this second draw, he draws a number greater than 21 he wins; otherwise, he loses. (a) Determine the sum that gives the probability that the player wins. (b) Write and run a line of \(\mathrm{R}\) code that computes the probability that the player wins. (c) Write an \(\mathrm{R}\) function that simulates the game and returns whether or not the player wins. (d) Do 10,000 simulations of your program in Part (c). Obtain the estimate and confidence interval, (1.4.7), for the probability that the player wins. Does your interval trap the true probability?

Short Answer

Expert verified
The sum that gives the probability that the player wins is approximately 2.926. The R code to compute this probability is: win_prob <- sum(4/(25-1:21))/20. An R function, simulate_game, is created to simulate the game and return whether the player wins. After running 10,000 simulations of the game, we get the estimate and confidence interval for the winning probability.

Step by step solution

01

Determine the probabilities of winning

(a) To determine the sum of probabilities, we must recognize that the player can only win if they draw 22, 23, 24 or 25 in the second round. These numbers can only be drawn if the first draw is 21 or less. Hence, we sum the probabilities of drawing 22, 23, 24 or 25 after having drawn x from the first set. This results in \(\sum_{x=1}^{21} \frac{4}{25-x} = \frac{4}{4} + \frac{4}{5} + \ldots + \frac{4}{24} = 1 + 0.8 + \ldots + 0.167 = 2.926\).
02

Writing R code to compute the winning probability

(b) After getting the sum, divide by the total number of possible outcomes (20) to get a probability. Write a line of R code for this: win_prob <- sum(4/(25-1:21))/20
03

Writing an R function to simulate the game

(c) Now we define an R function to simulate the game. The function draws a number from 1:20, then draws a second number from the first number to 25 and checks if it is greater than 21: simulate_game <- function() { first_draw <- sample(1:20, 1); second_draw <- sample(first_draw:25, 1); return(second_draw > 21) }
04

Running simulations to estimate and find the confidence interval of the winning probability

(d) Now that we have a game simulation function, we simulate the game 10,000 times and calculate the winning probability and the confidence interval. The R code for this is: simulation_results <- replicate(10000, simulate_game()); win_estimate <- sum(simulation_results)/10000; conf_interval <- qnorm(0.975)*sqrt(win_estimate*(1-win_estimate)/10000); paste('[', win_estimate - conf_interval, ', ', win_estimate + conf_interval, ']')

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a certain factory, machines I, II, and III are all producing springs of the same length. Machines I, II, and III produce \(1 \%, 4 \%\), and \(2 \%\) defective springs, respectively. Of the total production of springs in the factory, Machine I produces \(30 \%\), Machine II produces \(25 \%\), and Machine III produces \(45 \%\). (a) If one spring is selected at random from the total springs produced in a given day, determine the probability that it is defective. (b) Given that the selected spring is defective, find the conditional probability that it was produced by Machine II.

A mode of the distribution of a random variable \(X\) is a value of \(x\) that maximizes the pdf or pmf. If there is only one such \(x\), it is called the mode of the distribution. Find the mode of each of the following distributions: (a) \(p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(f(x)=12 x^{2}(1-x), 0

In an office there are two boxes of thumb drives: Box \(A_{1}\) contains seven 100 GB drives and three 500 GB drives, and box \(A_{2}\) contains two 100 GB drives and eight 500 GB drives. A person is handed a box at random with prior probabilities \(P\left(A_{1}\right)=\frac{2}{3}\) and \(P\left(A_{2}\right)=\frac{1}{3}\), possibly due to the boxes' respective locations. A drive is then selected at random and the event \(B\) occurs if it is a \(500 \mathrm{~GB}\) drive. Using an equally likely assumption for each drive in the selected box, compute \(P\left(A_{1} \mid B\right)\) and \(P\left(A_{2} \mid B\right)\)

A positive integer from one to six is to be chosen by casting a die. Thus the elements \(c\) of the sample space \(\mathcal{C}\) are \(1,2,3,4,5,6\). Suppose \(C_{1}=\\{1,2,3,4\\}\) and \(C_{2}=\\{3,4,5,6\\} .\) If the probability set function \(P\) assigns a probability of \(\frac{1}{6}\) to each of the elements of \(\mathcal{C}\), compute \(P\left(C_{1}\right), P\left(C_{2}\right), P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right)\).

Let \(X\) have the \(\operatorname{pmf} p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. Find the pmf of \(Y=X^{3}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free